In mathematics, algebras A, B over a field k inside some field extension Ω of k are said to be linearly disjoint over k if the following equivalent conditions are met:
(i) The map A ⊗ k B → A B induced by ( x , y ) ↦ x y is injective.(ii) Any k-basis of A remains linearly independent over B.(iii) If u i , v j are k-bases for A, B, then the products u i v j are linearly independent over k.Note that, since every subalgebra of Ω is a domain, (i) implies A ⊗ k B is a domain (in particular reduced). Conversely if A and B are fields and either A or B is an algebraic extension of k and A ⊗ k B is a domain then it is a field and A and B are linearly disjoint. However there are examples where A ⊗ k B is a domain but A and B are not linearly disjoint: for example, A=B=k(t), the field of rational functions over k.
One also has: A, B are linearly disjoint over k if and only if subfields of Ω generated by A , B , resp. are linearly disjoint over k. (cf. tensor product of fields)
Suppose A, B are linearly disjoint over k. If A ′ ⊂ A , B ′ ⊂ B are subalgebras, then A ′ and B ′ are linearly disjoint over k. Conversely, if any finitely generated subalgebras of algebras A, B are linearly disjoint, then A, B are linearly disjoint (since the condition involves only finite sets of elements.)
